Question 1: Let be a set function defined for all sets in a -algebra with values in . Assume is countably additive over countable disjoint collections of sets in .
(a) Prove that if and are sets in with , then .
(b) Prove that if there is a set in for which , then
(c) Let be a countable collection of sets in . Prove that .
Solution: (a) Note and are disjoint. Hence . Since , we have as desired.
(b) If is the only set for which , then for any set we have . Note and are disjoint. So . It follows that .
If , then and are disjoint. So . Since , we have .
(c) If , then the result holds trivially. So we assume the right hand side is finite.
Let and when . Then . Note the sets in are mutually disjoint, and each So by assumption we have
Question 2: Prove that for any countable set , its Lebesgue outer measure .
Proof: Let be an enumeration of . Then for any , is an open cover of . Let . So by definition of Lebesgue outer measure, . Since can be arbitrarily small, we must have .
Question 3: Let be the set of irrational numbers in the interval . Prove that Lebegue outer measure of is , i.e. .
Proof: Since , . On the other hand, let be the set rational numbers. Then is countable. By Question 2, we have . So . Thus, .
Question 4: A set is called a set if it is a countable intersection of open sets. Show that for any bounded set , there is a set that .
Proof: Since is bounded, there is some such that the open set covers . Thus, . By definition of Lebesgue outer measure, for any natural number , there is an open cover and , such that . So we construct a sequence of such open sets . Now take and is a set. We claim .
To see this, observe that every . Hence and we have . On the other hand, for every . Thus for every . As a result, we must have .
Question 5: Let be the set of rationals in . Let be a finite collection of open intervals that covers . Prove that .
Proof: Without loss of generality, we may assume has these properties: (1) ; (2) they are mutually disjoint.
So if , then there is some open interval in such that is disjoint with any interval in , and contains a rational number since the set of rational numbers is dense in . This contradicts with the fact that is collection of .
Question 6: Prove that if , then .
Proof: Observe . Conversely, since , .
Question 7: Let and be bounded sets for which there is an such that for all and . Prove that .
Proof: It suffices to prove . To this end, we need to show for any collection that covers , we have .
Without loss of generality, we may assume every contains some points in or or both. Consider an interval . Let and . Note has these properties: (1) ; (2) is open; (3) no belongs to . has similar properties. Note . Since is bounded by , we may express as a finite union of open intervals . Similarly, . Hence .
Note the countable union of covers and countable union of covers . So .
Question 8: Show that if has finite measure and , then is the disjoint union of a finite number of measurable sets, each of which has measure at most .
Proof: Observe that since . Hence there is some such that . Now partition to sufficiently smaller intervals each of which has length at most .
Question 9: (Lebesgue) Let have finite outer measure. Show that is measurable if and only if for each open, bounded interval , .
Proof: If is measurable, by definition the conclusion is true. Conversely, assume it holds for any open, bounded interval . Since has finite outer measure, by definition of outer measure, for any there is an open set containing such that . To prove the final result, it suffices to show for this we must have . In fact if (1) holds, then by outer approximation will be measurable.
To prove (1), note can be expressed as a countable union of disjoint open intervals, i.e. . So . By assumption, each . So . Observe that . Hence . Hence . Similarly, . Hence . Thus, (1) holds.
Let be a measurable function on . We write and . So . Observe that are all non-negative measurable functions on . Recall that for a non-negative measurable function , we say is integrable over if is well defined and .
Here we define the Lebesgue integration for a generic measurable function . And we state its various properties without proofs.
Lemma 1: Let be a measurable function on . Then is integrable over if and only if are integrable over . Then we define
Theorem 2 (Linearity and Monotonicity): For any , we have
And if on , then
Theorem 3 (Additivity of integral over domains):
Lemma 4 (Comparison Test): Let be measurable functions on . If a.e. on and is integrable over , then is also integrable over .
Lebesgue Dominated Convergence
Theorem 5 (Lebesgue Dominated Convergence Theorem): Let be a sequence of measurable functions on and pointwise a.e. on . Let be a measurable function that is integrable over . If dominates in the sense that on for every , then is integrable over and
Proof: It is easily shown that is measurable and a.e. on . By comparison test, is integrable over . It remains to show the validility of exchanging limit and integration.
Observe the sequence converges to pointwise a.e., and and are non-negative for all . Thus, we can apply Fatou’s Lemma and have . By (1) we further have . Similarly, we consider the sequence and the function , and easily show that . By (2) and (3) we have the result as desired.
Theorem 6 (General Lebesgue Dominated Convergence Theorem): Let be a sequence of measurable functions on and pointwise a.e. on . Let be a sequence of non-negative measurable functions and pointwise a.e. on . Furthermore, assume . If for every , then is integrable over and
Proof: It is easily shown that a.e. on (otherwise, there will be some which is a contradiction). So by comparison test, is integrable over .
Next, similar to the proof in Lebesgue Dominated Convergence, we consider sequence and function . Observer for every and . By Fatou’s Lemma and some manipulation we have . Similarly, we consider sequence and function , and obtain .
Countable Additivity and Countinuity of Integration
Theorem 7 (Countable Additivity of Integration): Let be integrable over . Let , where are measurable and mutually disjoint. Then,
Proof: Let . Define . So pointwise and . Since is integrable over , by Lebesgue Dominated Covergence, . By additivity of integration over domains, we have . By (1) and (2), we have .
Theorem 8 (Continuity of Integration): Let be integrable over .
(a) If is an increasing sequence of measurable subsets of , then
(b) If is a decreasing sequence of measurable subsets of , then
Proof: (a) Let and for . Then is a sequence of disjoint measurable subsets in . By additivity of integration in Theorem 7, we have Observe . So by (2) we have . By (1) and (3) we prove part (a).
(b) Let for . Then is an increasing sequence of measurable subsets in . By the result of part (a), we have . Observe . Since is decreasing, . So by (4) and (5) we have .
Vitali Convergence Theorem
Lemma 9: Let has finite measure. Then for any , is a union of finitely many disjoint measurable sets, each of which has measure at most .
Proof: Since has finite measure, . So there is a such that and . Then choose sufficient large s.t. and break to intervals and each $E_i = E \cap [-N+(i-1)\delta, -N+i\delta), i=1,2,…,M-1E_M = E \cap [-N+(M-1)*\delta, N]\blacksquare$
Lemma 10: Let be a measurable function on . If is integrable over , then for any , there is a s.t. for any and , . Conversely, let . If for any there is a s.t. for any and , then is integrable over .
Proof: Without loss of generality we may assume is non-negative. Suppose . There by definition of integration, there is a non-negative function s.t. , is bounded, has finite support in and . Since is bounded, let be such a bound. So for any , we have So choose we have the result as desired.
Conversely, assume and there is a to meet the challenge of . By Lemma 9, where and are disjoint. So .
We say a family of measurable functions on is uniformly integrable if for any there is a such that for any and any with , .
Theorem 11 (Vitali Convergence Theorem): Let has finite measure. Let be uniformaly intergrable over . Then if pointwise a.e., then is integrable over and .
Proof: Let meet the challenge of . Since has finite measure, by Lemma 9, where and are disjoint, s.t. for every . So for every . So by Fatou’s Lemma, . Thus, is integrable over .
Now consider , where and are disjoint. For any , by Egoroff’s theorem, there is a such that and whenever we have . By uniformly integrability over , the meets the challenge of (you should know the trick by choosing some where meets the challenge of Egoroff’s theorem and meets the challenge of uniform integrability). So we have , and . By (1)(2)(3) we have when .
If has infinite measure, we need additional property for besides uniform integrability to justify the passage of limit under integral sign.
Lemma 12: Let be integrable over . Then for any there is a set of finite measure such that .
Proof: Since , by definition of the integral there is a non-negative function which is bounded and has finite support in s.t. . Let the finite support be and in . Then, .
We say the family of measurable functions tight over if for any there is a subset of finite measure s.t. for every .
Theorem 13 (General Vitali Convergence Theorem): Let be uniformaly intergrable and tight over . Then if pointwise a.e., then is integrable over and .
Proof: For any , since is tight over there is a subset of finite measure s.t. for every . By Fatou’s Lemma, . So is integrable over .
Note has finite measure. By Vitali Convergence Theorem, is integrable over . Thus is integrable over . Furthermore, there is some s.t. whenever , . So , whenever .
Before we move on, we prove two corollaries of Vitali Convergence Theorems.
Corollary 14: Let be a sequence of non-negative functions that converge to pointwise a.e. on , where . Suppose is integrable over for every . Then, , if and only if, is uniformly integrable over .
Proof: By Vitali Convergence Theorem, if is uniformly integrable over , then . Conversely, suppose . Then for any there is some s.t. when , . So for any , for . Observe that the finite set of functions is uniformly integrable over with a that meets the challenge. So choose , then for every when . Hence is uniformly integrable over .
Corollary 15: Let be a sequence of non-negative functions that converge to pointwise a.e. on . Suppose is integrable over for every . Then, , if and only if, is uniformly integrable and tight over .
Proof: Note it is possible that . By General Vitali Convergence theorem, if is uniformly integrable and tight over , then .
Conversely, suppose . For any , there is some s.t. when . Since every is integrable over , by Lemma 12 there is some of finite measure s.t. for . Thus, take . So , and is true for every . Hence is tight over .
To prove uniformly integrability over , observe that for any the above statement shows that there is a subset of finite measure s.t. for every . Since has finite measure, by Corollary 14 is uniformly integrable over . So there is a s.t. for any with , . Now consider any general subset with . If , then . If , then . If both and are non-empty, then .
Converge in Measure
We say that a sequence of measurable functions defined on converges to in measure, if for any , . That is, the set of “bad” points that do not converge is becoming less and less as grows. Note that this “bad” set can keep changing. Therefore, convergence in measure does not imply convergence pointwise. The converse is true when has finite measure.
Theorem 16: Let has finite measure. Let converges to pointwise a.e. on . Then converges to in measure.
Proof: By Egoroff’s theorem, for any there is a measurable subset s.t. and converges to uniformly. So for any , there is some s.t. when when for . Therefore, , when . So when we fix , for any there is some that meets the challenge. Hence converges in measure.
Though convergence in measure does not imply convergence pointwise, but there is a sub-sequence convergence in pointwise.
Theorem 17: Let converges to in measure on . Then there is a sub sequence that converges to pointwise a.e.
Proof: Observe that there is a strictly increasing sequence of natural numbers s.t. , when . Let . Then . By Borel-Centalli Lemma, almost all belongs to finitely many . It means almost all satisfies that there is some s.t. when , . This is equivalent to the statement that for almost any , the sequence converges to . .
Corollary 18: Let be a sequence of non-negative functions on . Suppose is integrable over for every . Then, , if and only if, converges to in measure, is uniformly integrable and tight over .
Proof: Suppose , by Corollary 15 is uniformly integrable and tight over . Note here we don’t need any condition of convergence of to justify uniform integrability and tightness. To further prove converges to in measure, we apply Chebyshev’s inequality as for a . Since , for any there is some s.t. when , . Hence by (1) . So , and in measure.
Conversely, suppose converges to in measure, is uniformly integrable and tight over . Assume the conclusion does not hold. Then for any there is a sub-sequence s.t. . Note also converges to in measure. By Theorem 17, there is a further sub-sequence that converges to pointwise a.e. By General Vitalli Convergence, , contradicting the assumption that .
We want to study the notion of integration for a generic measurable function on . Here is general in terms of
takes value in , but can be unbounded.
The domain , on which is defined, is measurable but can have infinite measure
As in Royden’s book, we will establish the integration of a general case step by step:
Define Lebesgue integration for simple functions on any measurable set
Define Lebesgue integration for bounded functions on with finite measure
Define Lebesgue integration for non-negative functions on with any measure (can be infinite)
Define Lebesuge integration for general cases
For each scenario, we will:
establish linearity and monoticity
establish additivity of integral over domains
study convergence: under what conditions we can exchange and , i.e. if in some sense, when can we have
Also note that in general cases, the integral can be . Hence we are interested in cases where . We would call this case as “ is integrable over E“.
Lebesgue Integration for simple functions
In this note, we may skip the proofs of some simple lemmas and theorems.
Let be a simple functions on with finite measure, and express as , where takes the value in the measurable subset .
Then the Lebesgue integral of over is defined as
Theorem 1 (Linearity): Let be simple functions on with finite measure. For any , Proof: It suffices to prove (a) , and (b) .
For (a), notice that . Therefore, . This proves (a).
For (b), assume and . Let and . Then let . It is easily shown that only takes values in . For , define . It is easily shown that is measurable, and for any , and are disjoint. Thus, can be rewritten as , and . Hence
Theorem 2 (Monotonicity): Let be simple functions on . If on , then .
Proof: Since , . By linearity in Theorem 1, .
is bounded and has finite measure
Let be a bounded function defined on a measurable set with finite measure. is bounded, i.e. there is some s.t. on .
We define its upper Lebesgue integral over as We will often abbreviate the above to . We defined its lower Lebesgue integral over as Similary, we abbreviate the above to
You can easilly prove that
Lemma 3:
We say the Lebesgue integral of exists if . We write its integral as and have
Theorem 4: Let be a bounded function over a measurable set with finite measure. If is measurable, then is integrable over , i.e. exists.
Proof: By Simple Approximation Lemma (Lemma 3 in Note LIR3), for any there exist simple functions s.t. for all , and . Hence, . The above inequality holds for arbitrarily large . Therefore, we must have .
Theorem 5 (Linearity): Let be bounded measurable functions on with a finite measure. For any , .
Proof: Since are bounded and measurable, is also bounded and measurable (Theorem 7 in Note LIR2). Hence is integrable over by Theorem 4.
First we prove . If , observe that , and . Since , we have , which leads to .
If , it suffices to prove . By similar arguements, we can see that . The left hand side of (1) equals to . By (1) and (2) we obtain . This completes the proof of for any . (The case is trivial.)
Next, we proceed to prove . Observe, for arbitrary simple functions and , . By arbitrariness of , we must have . Similarly we have . By (3) and (4) we establish .
Corollary 6 (Monotonicity): Let be bounded measurable functions on with a finite measure. If on all of , then .
Proof: Observe . Then is a non-negative measurable function on . Then where on . Thus .
Corollary 7: If is bounded and integrable over a finite measure , then
Proof: is bounded and integrable over . Observe . So by linearity and monotonicity, we have $$
\int_E |f| \leq \int_E f \leq \int_E |f| $$ , which leads to the desired result.
Corollary 8 (Additivity of Integral over Domains): Let be a bounded and measurable function on a set of finite measure . Let be a measurable subset. Then
Proof: Define function as on and on . Similarly, define on and on . Then it is easily shown that are both measurable on . Observe , and . So by linearity we have .
Corollary 10: Let be a bounded and measurable function on a set of finite measure . If , then .
Proof: For any , its integral .
Theorem 9 (Uniformly Convergent Sequence): Let be a sequence of bounded and measurable functions on a set of finite measure . If uniformly, then
Proof: For any , since uniformly, there exists some s.t. . So .
If pointwise a.e. on , then we need some extra conditions to bound .
Theorem 10: (Pointwise Convergent and Uniformly Boundeded Sequence): Let be a sequence of bounded and measurable functions on a set of finite measure . If pointwise a.e. on and in addition is uniformly bounded, i.e. there is some s.t. on for all , then
Proof:
First by Theorem 1 in Note LIR 3, is measurable on and thus integrable over .
Let be the set that does not converge to , and . Note . If we were able to establish the result for , then , since by Corollary 10, and are all .
Hence we may assume pointwise on . It is easily shown that . So By Egoroff’s Theorem, for any , there is a and a closed set s.t. and whenever . Therefore, .
is non-negative and E has infinite measure
For a measurable function defined on a measurable set , we say has a finite support , if for , for all , and . We say vanishes outside .
If has a finite support of and is bounded in , then we define . The latter is well defined in the last section.
To define when and can have infinite measure, we use a set of non-negative bounded functions that have finite support to “approach” it. Formally,
When , we say is integrable over .
Theorem 11 (Equivalent Definition): If is a non-negative measurable function defined on a measurable set , then
Proof: Let , and let . Clearly , and thus . To prove the converse, we need to show for every , there exists some s.t. .
Note by Simple Approximation Theorem, there is a sequence in s.t. pointwise a.e. on and for all . Without loss of generality, we have pointwise on , where is the finite support of .
If is bounded on , then is uniformly bounded on . So
If is unbounded on , then let and . So for .
Since is bounded, is uniformly bounded on . Hence
On the other hand, is bounded on hence . So there is some such that and
Note is increasing and . By (1) and (2) we can choose sufficiently large such that .
Theorem 12 (Lineraity and Monotonicity): Let be non-negative measurable functions over a measurable set . For any ,
If for all , then
Proof: First let’s prove . For short, let’s write and . We want to show . By linearity of integral of bounded functions, for any we take . Then . So we can easily show .
Conversely, for any s.t. , we take . Then . Thus . So we have established .
Next we prove . Consider any with finite support , and any with finite support . Take . Then vanishes outside and we can easily show that . Clearly on . Hence we have . Conversely, consider a with finite support . Let s.t. on . Let . So on . Now take and on and on . So on and on . Also on . Thus, . This proves .
By (1) and (2) we prove the linearity of integration.
Finally, assume . Then any would satisfy . By definition of the integral, the supremum of left-hand-side must be less than or equal to the supremum of the right-hand-side. Hence .
Corollary 13 (Additivity of Integral over Domains): Let be a non-negative measurable function on . Let be a measurable subset in . Then
Proof: Define on and on . And define on and on . Then on . By linearity, we have . Note the finite support of is . So it is not difficult to show . Similarily we have .
Theorem 14 (Fatou’s Lemma): Let be a sequence of measurable functions that converges to on pointwise a.e. Then is measurable on and
Proof: Without loss of generality, we may assume pointwise. We need to show for any bounded non-negative mesurable functions with finite support, .
To this end, let be the finite support of and let on . It is easil shown that that pointwise. Since is bounded, is bounded for any . Hence by convergence theorem, we have . On the other hand, . By the definition of , we have . Therefore, .
Theorem 15 (Monotone Convergence Theorem): et be an increasing sequence of measurable functions that converges to on pointwise a.e. Then
Proof: By Fatou’s Lemma, we get . It reamins to show . Since is increasing, . So . Thus we have (1) as desired.
Theorem 16 (Chebyshev’s Inequality) For any ,
Proof: Let . We only prove the case when . Let . Then is a bounded non-negative functon with finite support. Clearly on . So .
Corollary 17: if and only if a.e. on
Proof: If a.e. on , then any bounded non-negative with finite support must satisfy a.e. on . Then and thus .
Conversely, let . By Chebyshev’s inequality, for every natural number . Note . So , i.e. a.e. on .
Let be a sequence of functions defined on a measurable set . There are some different types of convergence.
Converge uniformly: we say converges to uniformly on , denoted by , if for any there is a s.t. whenever , for any .
Converge point wise: we say converges to pointwise, denoted by , if for any , . We say converges to pointwise almost everywhere, denoted by , if on and
Converge in measure: we say converges to in measure, denoted by , if for any and there is a such that whenever we have
The notion of “convergence in measure” is that the points () whose do not converge to is getting smaller and smaller. But this does not gaurantee that every point will converge to as in “converge pointwise” or even in “converge pointwise almost everywhere”. In probability theory, “converge in measure” is also known as “converge in probability”, and “converge pointwise (almost everywhere)” is also known as “converge almost surely”.
In probability theory, “converge pointwise (almost everywhere)” can also be denoted by , while “converge in probability” can be denoted by the statement that for ,
“Converge uniformly” is the strongest version of convergence among the three, as it only requires convergence for each point but also requires they “converge at the same speed”, i.e. after a certain threshold , all must be very close () to .
Now we see how convergence preserves measurability.
Theorem 1: is a sequence of measurable functions on . If pointwise a.e., then is measurable on .
Proof: For any , let’s consider the set . Observe , which is measurable.
Approximation
, defined on a measurable set , is said to be simple if it only takes a finite number of values and each is measurable. So we often write as , where is an indicator function, i.e. if is true and 0 otherwise.
For a function, we can use a sequence of simple functions to approximate it. Formally, we have the following theorem.
Theorem 2 (Simple Approximation Theorem): Let be a function defined on . Then is measurable, if and only if, there is a sequence of simple functions such that for any and pointwise. If is non-negative, we may choose as increasing.
To prove Theorem 2, we first prove a lemma.
Lemma 3 (Simple Approximation Lemma): Let be a bounded measurable function on , i.e. there is an such that for any . Then for any , there are simple functions such that , and
Proof: Since is bounded by , for partition the interval to , where , and for each , .
Let . Then is measurable. Define and .
Proof of Theorem 2: By Theorem 1, if the sequence of simple functions converges to pointwise, then is measurable. It remains to prove the converse.
Assume is non-negative. Note can be unbounded. Let and
(for some Markdown to show correctly, we have if and if ).
So by Lemma 3, there are defined on such that when for any and . We claim that is the increasing sequence of simple functions as desired.
To see this, consider any . Note is increasing with . So there is a such that whenever . For any there is an such that . So whenever . Take . So for , there exists such that when , we have (1) and (2) . Hence pointwise. This proves the cases where is non-negative.
Assume is non-positive, then is non-negative. By above arguments, there is a sequence s.t. and point wise. So we take and it is easily shown that it meets the requirement.
Finally, let be arbitrary. Then . The corresponding sequences are . Take . It is easily shown that pointwise and .
Egoroff’s Theorem
Egoroff’s theorem is another example of Littlewood’s principles, i.e. every pointwise convergence is “almost” uniform convergence.
Formally,
Theorem 4 (Egoroff’s Theorem): Let be a measurable set with finite measure. Let be a sequence of measurable functions on . pointwise on . Then for any there is a closed set and such that uniformly.
Again, we prove a lemma first.
Lemma 5: Let satisfy the conditions stated in Egoroff’s Theorem, then for any there exists such that and for any whenever .
Proof: Let . Since pointwise, for any there is some such that whenever . Now we “group” these into different sets as follows.
Let . Then is increasing and . So by continuity of Lebesgue measure, we have . Since , there is some such that . Let be , and the constructed are what we desire.
Proof of Egoroff’s Theorem: For any , by Lemma5, take and . Then there are some and s.t. and , for any whenever .
Note is measurable. So take and is also a measurable set. Observe that uniformly on , and . Since is measurable, by inner approximation (Theorem 11 in note LIR1), there is a closed set s.t. . Therefore uniformly on the closed set and .
Let be a measurable set and let a extended-value function be defined on , i.e. .
We say is a measurable function if its inverse image of any Borel set is measurable. Formally,
Theorem 1: The following statements are equivalent. And if holds any of the following property, we say is a measurable function.
For any , the set is measurable.
For any , the set is measurable.
For any , the set is measurable.
For any , the set is measurable.
Proof: Assume Item 1 holds. Let . So . Each is measurable, so Item 2 holds.
Assume Item 2 holds. The set in Item 3 is a complement of the set in Item 2. Hence Item 3 holds.
Assume Item 3 holds. By taking similar approach in proving Item 1 implies Item 2, we obtain the fact that Item 4 holds.
Assume Item 4 holds, the set in Item 1 is a complement of the set in Item 4. So Item 1 holds.
Corollary 2: If is a measurable function defined on the measurable set , then for any , the set is measurable.
Proof: If , note . The two sets on the right hand side are measurable, so is the one on left hand side.
If , consider . Then , and it is measurable. Similarly, the conclusion holds when .
Corollary 3: is a measurable function defined on the measurable set , if and only if, for any open set , is measurable.
Proof: For any , is an open set. So is measurable. Hence is a measurable function.
Coversely, suppose is measurable and consider an open set . It can be expressed as a union of disjoint open intervals, i.e. . Let and . Then and are both measurable sets, and is also measurable.
So is measurable.
Theorem 4 (Continuity implies measurability): Let be a continuous function on a measurable set . Then is measurable on .
Proof: If is continuous, for any open set we have where is open. Thus is measurable. By Corollary 3, is measurable.
Theorem 5: Let be a monotonic function on an interval . Then is measurable.
Proof: Assume is increasing in .
Suppose is an open interval where . For any , if then the set is measurable. If , the set is measurable. If , the set is measurable.
Suppose is a left-closed-right-open interval , where . For any , if then is measurable. If , is measurable. If , it is an empty set.
Similar statements can be applied to the cases and .
If is decreasing, simiarly we can show that it is also measurable.
Next, we wonder if is measurable, and functions have certain relationships with , how about their measurability?
Theorem 6: is a function on .
If is measurable and almost everywhere on , then is measurable on .
Let be a measurable subset of . Then is measurable on if and only if is measurable on both and .
Proof: For statement (1), let be the set that and on . Then for any , the set , which is measurable. This proves statement (1).
For statement (2), for any let and . If is measurable on then is measurable. Thus both and are measurable, which implies is measurable on both and . Conversely, if is measurable on and , the sets and are measurable. So is measurable.
We shall now only consider real-valued functions, i.e. , from a practical point of view and to simply a lot of proofs.
Theorem 7: are measurable functions on .
For any and , is measurable on .
is measurable on .
Proof: For statement (1), it suffices to prove (1a) is measurable, and (1b) is measurable.
For any , the set is measurable. This proves (1a).
Now consider the set . Then . Note the set of rationals is dense in . So there is a s.t. . Thus
Each component under the union is measurable and is countable. So the set on the left hand side of (1) is also measurable. This proves (1b).
For statement (2), we consider . Since we have establish statement (1), it remains to prove is measurable.
Consider the set . If , is measurable. , , which is also measurable.
Now we consider composition.
Theorem 8: Let is a measurable function on and is continuous on . Then the composition is measurable on .
Proof: Consider any open set . Then. . Since is continuous on , is open. Thus by Corollary 3, is measurable. Again by Corollary 3, is a measurable function.
Theorem 9: and are measurable on . Then
is measurable.
is measurable.
is measurable.
is measurable.
is measurable.
Proof: The set is measurable. This proves statement (1).
Similarly, the set is measurable. This proves statement (2).
The set if , and if . In either case, is measurable. This proves statement (3).
The set if and if . In either case, is measurable. This proves statement (4).
For statement (5), note . By (3), (4) and Theorem 8 (linearity preserves measurability), is measurable.